2025 AMC 12A Problems/Problem 10

Problem

In the figure shown below, major arc $\widehat{AD}$ and minor arc $\widehat{BC}$ have the same center, $O$ . Also, $A$ lies between $O$ and $B$ , and $D$ lies between $O$ and $C$ . Major arc $\widehat{AD}$ , minor arc $\widehat{BC}$ , and each of the two segments $\overline{AB}$ and $\overline{CD}$ have length $2\pi$ . What is the distance from $O$ to $A$ ?

$[asy] usepackage("mathptmx"); size(6cm); defaultpen(linewidth(0.7)); real r = 1 - pi + sqrt(pi^2 + 1); pair O = (0, 0), A = r * dir(-30), B = (r + 2pi) * dir(-30), C = (r + 2pi) * dir(30), D = r * dir(30); draw(arc(O, r, 30, 330)); draw(arc(O, r + 2pi, -30, 30)); draw(O--A, dashed); draw(O--D, dashed); draw(A--B); draw(C--D); dot("$O$", O, dir(135)); dot("$A$", A, dir(270)); dot("$B$", B, dir(0)); dot("$C$", C, dir(0)); dot("$D$", D, 1.5 * dir(70)); [/asy]$

$\textbf{(A)}~1 \qquad \textbf{(B)}~1 - \pi + \sqrt{\pi^{2} + 1} \qquad \textbf{(C)}~\frac{\pi}{2} \qquad \textbf{(D)}~\frac{\sqrt{\pi^{2} + 1}}{2} \qquad \textbf{(E)}~2$

Solution 1 (Simple)

The ratio between the radius and the arc length is constant. For the inner circle, the radius, which we will denote as $r$ , has a corresponding arc length of $2\pi r - 2\pi$ (the circumference minus the major arc length). For the outer circle, the radius, which is $r + 2\pi$ , has a corresponding arc length of $2\pi$ . We therefore write the equation

$\frac{r}{2\pi r - 2\pi} = \frac{r+2\pi}{2\pi},$ which simplifies to $r = (r+2\pi)(r-1)$ $0 = r^2 + (2\pi-2)r - 2\pi.$ Applying the Quadratic Formula, we get that $r =\boxed{\textbf{(B)} 1 - \pi + \sqrt{1 +\pi^2}}$

~lprado

Solution 2

Let the length of $OA=r$ , which is the radius of the smaller circle. Then, the radius of the larger circle, $OB$ , is equal to $r+2\pi$ . Indeed, we know that the length of major arc $\widehat{AD}=2\pi$ and the length of minor arc $\widehat{BC}=2\pi$ . So, using the formula for length of an arc formed by the central angle $\angle COB$ , which we denote as $\theta$ , we have that:

$\widehat{AD}=2\pi=(2\pi-\theta) \cdot r,$ $\widehat{BC}=2\pi=\theta \cdot (2\pi+r).$

Expanding, we have $\theta \cdot r +2 \pi \theta = 2\pi,$ $2 \pi r-\theta r=2\pi,$ and by adding the two equations we have that

$2 \pi \theta + 2 \pi r =4\pi \implies \theta + r=2.$

Indeed, the question is asking for us to solve for $r$ , and so we use $\theta = 2-r$ back into our original equation to solve:

$(2 \pi - \theta) \cdot r = (2 \pi - 2 +r) \cdot r = (2 \pi -2)r +r^2= 2\pi \implies$ $r^2+(2 \pi -2)r - 2 \pi =0.$

Using the quadratic formula, we have that $r=\frac{2-2\pi \pm \sqrt{(2 \pi -2)^2 - 4(1)(-2\pi)}}{2}=\frac{2-2\pi \pm \sqrt{4 \pi^2 -8 \pi +4+8 \pi}}{2}= \frac{2-2\pi \pm 2 \sqrt{1+ \pi^2}}{2}= 1- \pi \pm \sqrt{1+\pi^2}.$

Since length must be positive, we consider only the positive root, and so the answer is $\boxed{\textbf{(B)} 1 - \pi + \sqrt{1 +\pi^2}}$ .

~e_is_2.71828

Video Solution by Power Solve

https://www.youtube.com/watch?v=BstSjBL0Bz8

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

2025 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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